If $f(x)$ is strictly increasing in an interval $[a,b]$ then what can we say about another function $g(x)=(f(x))^2$ in the same interval? I want to know whether $g(x)$ will be increasing, decreasing or whether it's nature cannot be predicted in $[a,b]$.
My attempt:
If $f(x)$ is strictly increasing in $[a,b]$ then we can say that $f'(x)>0$ in the interval $[a,b]$. Now, $g'(x) = 2*f'(x$) which will also be greater than $0$ in the interval $[a,b]$ as $f(x)$ is greater than zero in that interval. So, $g(x)$ should be strictly increasing too in the interval $[a,b]$.
But in the textbook, it is given that the nature of $g(x)$ cannot be predicted in the interval $[a,b]$ and I don't see any way to conclude that.
Any help would be appreciated.  
 A: An important mistake :
IF $f$ is a continuous function and also differentiable with $f'>0$ in some interval, truly then $f$ will be strictly increasing in this interval.
BUT IF $f$ is a strictly increasing function on an interval, this does not necessarily mean that $f'>0$ in that interval as well.
Keeping this in mind, you're given that $f$ is strictly increasing in $[a,b]$. This means (by the definition of when a continuous function is strictly increasing) that $\forall x_1,x_2\in [a,b] \space \text{with} \space x_1<x_2$ it is : $$f(x_1)<f(x_2)$$
This definition though cannot be checked for $f^2(x)$, as if $x_1$ is negative and $x_2$ is positive, then it would be $f^2(x_1) > f^2(x_2)$ if $|x_1| > |x_2|$ while $x_1<x_2$. Thus no conclusion can be made about $g(x)=f^2(x)$.
For some counter-examples, you can check Jose Carlos Santos answer down below.
A: It cannot be predicted. Take $f_1,f_2,f_3\colon[0,1]\longrightarrow\mathbb R$ defined by $f_1(x)=x-1$, $f_2(x)=x-\frac12$, and $f_3(x)=x$. All of them are increasing, but ${f_1}^2$ is decreasing, ${f_2}^2$ is neither increasing nor decreasing, and ${f_3}^2$ is increasing.
